PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Soil SolutionSamplingSoluble ComplexesSpeciationThermodynamic Stability ConstantExtraction MethodsCollect drainage water in situReaction with collection vesselMust be at or near saturationHigh variabilityDisplace with an immiscible liquidAs by F3Cl3C2 in centrifuge (ρ = 1.58 g cm-3)Displace using air pressure (positive or vacuum)Reaction with filterMust be at or near saturationDisplace using centrifugal forceGenerally, the extract cannot be identical to the true soil solutionCriticize this approach.Particle density = 2.62 g cm-3 and bulk density = 1.32 g cm-3so that porosity = 0.50. What is the composition of saturated soil solution?Equilibrate 10 g water + 1 g soil, centrifuge and analyze 5 g water + 1 g soil 5 g water + 5 g soil10 5 1 0.37Empirically model Ca2+, Mg2+, …, SO42-, … and extrapolate.Soluble ComplexesComplex consists of a molecular unit (e.g., ion) as a center to which other units are attracted to form a close associationExamples include Si(OH)4 and Al(OH)2+ with Si4+ and Al3+ as the central unit and OH- as ligandsIf two or more functional groups of a ligand are coordinated to a central metal, complex is called a chelateIf central unit and ligands are in direct contact, complex is inner-sphereIf one or more H2O in between, complex is outer-sphereIf ligands are H2Os, complex is solvation complex (e.g., Ca(H2O)62+)What would be orientation of H2Os?Mg2+(aq) + SO42-(aq) = MgSO4(aq)Given that –COO- tends to form water bridges to soil minerals viadivalent cations adsorbed onto mineral surfaces, is the MgSO4 complex inner- or outer-sphere?Kinetics of complex formation are fastAssume described by rate of disappearance of the metalas in the rate of its concentration decrease,-d[Mg2+] / dt = Rf – Rrwhere forward and reverse rates are affected by the temperature, pressure and composition of the solutionFurther, typically,-d[Mg2+] / dt = kf [Mg2+]α[SO42-]β – kr[MgSO4]γαth order with respect to Mg2+If tracked formation under conditions of excess SO42-, could determine α-d[A] / dt = kf [A]α0th- d[A] / dt = kf and [A] = -kft + [Ao]1st- d[A] / dt = kf[A] and ln[A] = -kft + ln[Ao]2nd- d[A] / dt = kf[A]2and 1/[A] = kft + 1/[Ao]What if-d[Mg2+] / dt = kf [Mg2+][SO42-] – kr[MgSO4]at equilibrium0 = kf [Mg2+][SO42-] – kr[MgSO4][MgSO4] / [Mg2+][SO42-] = kf / kr = cKsDo problem 2.Al3+ + F- = AlF2+-d [Al3+] / dt = kf [Al3+][F-] – kr [AlF2+]At early stage of reaction, ignore reverse rate and since initial concentrationsof Al3+ and F- are the same, [F-] = [Al3+], giving-d [Al3+] / dt = kf [Al3+]2or-d [Al3+] / [Al3+]2 = kf dt which integrates to1 / [Al3+] – 1 / [Al3+]0 = kf tNow what is the time when ½ of the Al3+ has reacted, i.e., the half life?1 / ½ [Al3+]0 – 1/ [Al3+]0 = kf t½ ort½ = (1 / kf) (1 / [Al3+]0) = 909 s @ pH 3.9, given kf = 110 M-1 s-1= 138 s @ pH 4.9, given kf = 726 M-1 s-1Speciation EquilibriaAssumption of fast complex formation and slow redox / precipitationAl speciation exampleLimit possible ligands to SO42-, F- and fulvic acid (L-)Set pH = 4.6 for which AlOH2+ is major hydroxide complexIgnore polymeric forms of aluminum[Al]T = [Al3+] + [AlOH2+] + [AlSO4+] + [AlF2+] + [AlL2+]Use conditional stability constants to express concentrations in terms of [Al3+], [OH-] or [H+] and concentrations of ligand species[AlOH2+] = cK1 [Al3+][OH-][AlSO4+] = cK2 [Al3+][SO42-][AlF2+] = cK3 [Al3+][F-][AlL2+] = cK4 [Al3+][L-] [Al]T = [Al3+] { 1 + cK1 [OH-] + cK2 [SO42-] + cK3 [F-] + cK4 [L-]}Similarly,[SO42-]T = [SO42-] { 1 + cK2 [Al3+]}[F-]T = [F-] { 1 + cK3 [Al3+]}[L-]T = [L-] { 1 + cK4 [Al3+]}Now proceed iteratively,Step 1 begin[Al3+] = [Al]T / { cK1 [OH-] + cK2 [SO42-] + cK3 [F-] + cK4 [L-]}where [OH-] is known from pH and concentration of other ligands are assumed equal to their known total concentrationConcentrations of ligands other than OH- then calculated, as with[SO42-] = [SO42-]T / { 1 + cK2 [Al3+]}Step 1 endUse revised concentrations of ligands to improve estimate of [Al3+], beginning Step 2Continue until convergence reached Change in estimated concentrations from Stepi to Stepi + 1 < arbitrary criterionLimitations to predicting speciationCompleteness –may ignore important reactions (redox and speciation)Insufficient data –do not have conditional stability constantsAnalytical methodology –short-comings as in failure to distinguish between monomeric / polymeric or dissolved / particulate formsAssumption of equilibrium –ignores kineticsField soils –conditions vary from those for which cKis determined; spatial / temporal variability, particularly mass inputs / outputsSpreadsheet calculation problem[Al]T = 0.000010 M[SO42-]T = 0.000050 M[F-]T = 0.000002[L-]T = 0.000010pH = 4.60cK1 = 109.00 M-1cK2 = 103.20cK3 = 107.00cK4 = 108.60Solve for all formsSee spreadsheet.Also, do problem 6.The distribution coefficients, αis, = [H2CO3] / [CO3T], etc., where[CO3T] = [H2CO3] + [HCO3-] + [CO32-] [1]Using the given CKSs, [HCO3-] = CK2 [H+] [CO3-2] but[CO32-] = [H2CO3] / (CK1 [H+]2) so[HCO3-] = (CK2 / CK1) ([H2CO3] / [H+])and substituting in [1] gives[CO3T] = [H2CO3] {1 + (CK2 / CK1) / [H+] + 1 / (CK1 [H+]2)}and using [H+] = 10-pH the distribution coefficient for H2CO3 isαH2CO3 = 1 / {1 + 10-6.4 10pH + 10-16.7 102pH}Proceeding similarly,αHCO3 = 1 / {106.4 10-pH + 1 + 10-10.3 10pH}αCO3 = 1 / {10-16.7 10-2pH + 1010.3 10-pH +1}Now, when is HCO3- dominant, i.e., αHCO3 ≥ 0.5?This is the case when {106.4 10-pH + 1 + 10-10.3 10pH} ≤ 2, no?This is almost the case when either pH = 6.4 or pH = 10.3 becauseat either pH, {106.4 10-pH + 1 + 10-10.3 10pH} is only slightly greater than 2, e.g.,{106.4 10-6.4 + 1 + 10-10.3 106.4} = 1 + 1 + 0.00013 = 2.00013Thus, HCO3 is (approximately) dominant at 6.4 ≤ pH ≤ 10.3The earlier [Al3+] speciation problem can be handled more efficiently. Express the concentrations of ligands in terms of [Al3+] and total concentration of each ligand to give,[Al3+] = [Al]T / { (cK1 KW / [H+]) + (cK2 [SO42-]T / {1 + cK2 [Al3+]}) + (cK3 [F-]T / {1 + cK3 [Al3+]}) + (cK4 [L-]T / {1 + cK4 [Al3+]})}and approximate a solution for [Al3+] (below). In turn, the equilibriumconcentrations of all species are known from the relations,[SO42-] = [SO42-]T / {1 + cK2